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compute the limit. if does not exist, explain: lim f(x)...
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\[\lim_{x \rightarrow \infty} 2x ^{2}-3x / 1-2x-3x ^{2}\]
multiply top and bottom by 1/x^2
so lim[ 2x^2/x^2 -3x / (2x^2) ]/ [ 1 /x^2 -2x/x^2 - 3x/x^2]
A type of L'Hopital application: infinity/infinity First derivation step: (4x-3) / (-2-6x) still infinity/infinity Second step: 4 / (-6) = -2/3 This is the solution. *
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